Find the positive value of a for which the eq | vedclass

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(B) Given,$a > 0$. $f(x) = 2 x^3 - 9 a x^2 + 12 a^2 x + 1$. Find the derivative: $f'(x) = 6 x^2 - 18 a x + 12 a^2$. Factor the derivative: $f'(x) = 6

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$2$

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(B) Given,$a > 0$. $f(x) = 2 x^3 - 9 a x^2 + 12 a^2 x + 1$. Find the derivative: $f'(x) = 6 x^2 - 18 a x + 12 a^2$. Factor the derivative: $f'(x) = 6 (x^2 - 3 a x + 2 a^2) = 6 (x - 2 a) (x - a)$. Setting $f'(x) = 0$ gives critical points $x = a$ and $x = 2 a$. Find the second derivative: $f''(x) = 6 (2 x - 3 a) = 12 x - 18 a$. Check for local maxima and minima: At $x = a$,$f''(a) = 6 (2 a - 3 a) = -6 a  0$),so $\alpha = a$ is the point of local maximum. At $x = 2 a$,$f''(2 a) = 6 (4 a - 3 a) = 6 a > 0$ (since $a > 0$),so $\beta = 2 a$ is the point of local minimum. Given the condition $2 \alpha + \beta = 8$,substitute $\alpha = a$ and $\beta = 2 a$: $2(a) + 2 a = 8$. $4 a = 8$. $a = 2$.

Find the positive value of $a$ for which the equality $2 \alpha + \beta = 8$ holds,where $\alpha$ and $\beta$ are the points of maximum and minimum,respectively,of the function $f(x) = 2 x^3 - 9 a x^2 + 12 a^2 x + 1$.

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